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\(\frac{24}{x^2+2x-8}-\frac{15}{x^2+2x-3}=2\)2
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ND
1
Y
26 tháng 4 2019
Đặt \(x^2+2x-3=t\) thì pt trở thành :
\(\frac{24}{t-5}-\frac{15}{t}=2\)
\(\Leftrightarrow\frac{24t-15\left(t-5\right)}{t\left(t-5\right)}=2\)
\(\Leftrightarrow9t+75=2\left(t^2-5t\right)\)
\(\Leftrightarrow9t+75=2t^2-10t\)
\(\Leftrightarrow2t^2-19t-75=0\)
\(\Leftrightarrow2\left(t^2-\frac{19}{2}t-\frac{75}{2}\right)=0\)
\(\Leftrightarrow\left(t-\frac{25}{2}\right)\left(t+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\frac{25}{2}\\t=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+2x-3=\frac{25}{2}\\x^2+2x-3=-3\end{matrix}\right.\)
1 tháng 7 2019
\(\frac{\left(x+2\right)^2}{8}-2\left(2x+1\right)=25+\frac{\left(x-2\right)^2}{8}\)
\(\Leftrightarrow\frac{\left(x+2\right)^2}{8}-\frac{16\left(2x+1\right)}{8}=\frac{200}{8}+\frac{\left(x-2\right)^2}{8}\)
\(\Leftrightarrow\left(x+2\right)^2-32x-16=200+\left(x-2\right)^2\)
\(\Leftrightarrow x^2+4x+4-32x-16-200=x^2-4x+4\)
\(\Leftrightarrow x^2-28x-212-x^2+4x-4=0\)
\(\Leftrightarrow-24x=216\)
\(\Leftrightarrow x=-9\)
1 tháng 7 2019
TL:
a)
\(\frac{\left(x+2\right)^2}{8}-\frac{16\left(2x+1\right)}{8}=\frac{200+\left(x-2\right)^2}{8}\)
\(\frac{x^2+4x+4-32x-16}{8}=\frac{200+x^2-4x+4}{8}\)
\(x^2-28x-12-200-x^2+4x-4=0\)
\(-24x-216=0\)
\(-24x=216\)
\(x=-9\)
Vậy x=-9
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 10 2019
ĐKXĐ:
a/ \(\left\{{}\begin{matrix}x^2+2x-8\ne0\\x^2+2x-3\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne-4\\x\ne2\\x\ne1\\x\ne-3\end{matrix}\right.\)
b/ \(x^2-2x+3\ge0\Rightarrow x\in R\)
3 tháng 8 2020
\(5X\left(X-2020\right)+X=2020\)
\(\Leftrightarrow5X^2-10100X+X=2020\)
\(\Leftrightarrow5X^2-10099X=2020\)
\(\Leftrightarrow5X^2-10099X-2020=0\)
\(\Leftrightarrow5X^2-10100X+x-2020=0\)
\(\Leftrightarrow5X\left(X-2020\right)+X-2020=0\)
\(\Leftrightarrow\left(X-2020\right)\left(5X+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-\frac{1}{5}\end{cases}}\)
3 tháng 8 2020
\(4\left(x-5\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)\right]^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)-2x-1\right]\left[2\left(x-5\right)+2x+1\right]=0\)
\(\Leftrightarrow\left(2x-10-2x-1\right)\left(2x-10+2x+1\right)=0\)
\(\Leftrightarrow-11\left(4x-9\right)=0\)
\(\Leftrightarrow x=\frac{9}{4}\)
5 tháng 3 2022
c: \(=\dfrac{1}{3x-2}-\dfrac{4}{3x+2}+\dfrac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-12x+8+3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{-6x+4}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-2}{3x+2}\)
d: \(=\dfrac{x^2-4-x^2+10}{x+2}=\dfrac{6}{x+2}\)
e: \(=\dfrac{1}{2\left(x-y\right)}-\dfrac{1}{2\left(x+y\right)}-\dfrac{y}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{x+y-x+y-2y}{2\left(x-y\right)\left(x+y\right)}=\dfrac{0}{2\left(x-y\right)\left(x+y\right)}=0\)
KT
1
IA
18 tháng 3 2020
\(\Leftrightarrow\frac{6x^2+3}{24}-\frac{10x-4}{24}=\frac{6x^2-6}{24}-\frac{4x-12}{24}\)
\(\Leftrightarrow\frac{6x^2+3-10x+4}{24}=\frac{6x^2-6-4x+12}{24}\)
\(\Leftrightarrow6x^2-10x+7=6x^2-4x+6\)
\(\Leftrightarrow-6x+1=0\)
\(\Rightarrow-6x=-1\)
\(\Leftrightarrow x=\frac{1}{6}\)
Vậy ...
21 tháng 9 2019
a) \(\frac{2x}{x+2}+\frac{x+2}{2x}=2\)
\(\Leftrightarrow4x^2+\left(x+2\right)^2=4x\left(x+2\right)\)
\(\Leftrightarrow5x^2+4x+4=4x^2+8x\)
\(\Leftrightarrow5x^2+4x+4-4x^2-8x=0\)
\(\Leftrightarrow x^2-4x+4=0\)
\(\Leftrightarrow x^2-2.x.2+2^2=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
\(\Rightarrow x=2\)
BẰng 2 chứ ko phải 22
\(ĐKXĐ:\hept{\begin{cases}x\ne2;x\ne-4\\x\ne1;x\ne-3\end{cases}}\)
\(\frac{24}{x^2+2x-8}-\frac{15}{x^2+2x-3}=2\)
\(\Leftrightarrow\frac{24}{\left(x-2\right)\left(x+4\right)}-\frac{15}{\left(x-1\right)\left(x+3\right)}-2=0\)
\(\Leftrightarrow24\left(x^2+2x-3\right)-15\left(x^2+2x-8\right)-2\left(x^2+2x-8\right)\left(x^2+2x-3\right)=0\)
\(\Leftrightarrow24x^2+48x-72-15x^2-30x+120-2\left(x^4+4x^3-7x^2-22x+24\right)=0\)
\(\Leftrightarrow9x^2+18x+48-2x^4-8x^3+14x^2+44x-48=0\)
\(\Leftrightarrow-2x^4-8x^3+23x^2+62x=0\)
\(\Leftrightarrow-x\left(2x^3+8x^2-23x-62\right)=0\)
\(\Leftrightarrow-x\left(2x^3+4x^2+4x^2+8x-31x-62\right)=0\)
\(\Leftrightarrow x\left[2x^2\left(x+2\right)+4x\left(x+2\right)-31\left(x+2\right)\right]=0\)
\(\Leftrightarrow x\left(x+2\right)\left(2x^2+4x-31\right)=0\)
\(\Leftrightarrow\)\(x=0\)
hoặc \(x=-2\)
hoặc \(2\left(x+1\right)^2-33=0\)
\(\Leftrightarrow\)\(x=0\)(tm)
hoặc \(x=-2\)(tm)
hoặc \(x=-\frac{2\pm\sqrt{66}}{2}\)(tm)
Vậy tập nghiệm của phương trình là \(S=\left\{0;-2;-\frac{2\pm\sqrt{66}}{2}\right\}\)